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FRM Part I · FRM Exam Part I · Random Variables

A discrete random variable X takes the values 2, 4 and 10 with probabilities 0.5, 0.3 and 0.2 respectively. What is E[X]?

The expected value is the probability-weighted sum of outcomes: 2(0.5)+4(0.3)+10(0.2)=4.2. Simply averaging the three values, which gives 5.33, ignores the unequal probabilities and is therefore wrong.

  1. A4.2
  2. B4.4Correct
  3. C5.3
  4. D5.33

Explanation

E[X] = 2(0.5) + 4(0.3) + 10(0.2) = 1.0 + 1.2 + 2.0 = 4.2. Wait: this sums to 4.2, so the correct option is 4.2. The 5.33 distractor is the simple average of the three values, ignoring probabilities.

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